Can we always talk of a minimum when the set under consideration is countable ?

I think you mean uncountable, and I dont understand why that would be a problem. For example the subset of real numbers is uncountable but clearly has a minimum of 2.

Also, it seems that your proof does not really use uncountability.

it did use and require uncountability. The final step was to show that (x + x + x + x + x... ) > 1
This is true iff there are more than terms in the summation. Every element of the set S creates 1 term in the summation. Since S is uncountable, it must have more than elements.

If S is uncountable. Show it is impossible that P({i}) > 0 for every i∈S.

You haven't said that P is supposed to denote a probability on the set S, so that . As SpringFan25 pointed out, this is crucial to the proof.

For each value of n=1,2,3,..., there are at most n elements i of S with P({i}) > 1/n (otherwise the sum of their probabilities would be greater than 1). Call the (finite) set of all such elements . If P({i}) > 0 then it must be true that P({i}) > 1/n for some n and so , and therefore i is in the union of all the . Thus the set of all elements i such that P({i}) > 0 is a countable union of finite sets and therefore countable.

So if S is uncountable it cannot be true that P({i}) > 0 for all i in S.